Question

On a certain​ route, an airline carries 8000 passengers per​ month, each paying ​$70.A market survey indicates that for each​ $1 increase in the ticket​ price, the airline will lose 50 passengers. Find the ticket price that will maximize the​ airline's monthly revenue for the route. What is the maximum monthly​ revenue?

Answer 1

The maximum revenue is $661,250.

Explanation:

The total revenue (R) is the product between the number of passengers (N) and the price of each ticket (P), so the inicial revenue is:

R = N * P = 8000 * 70 = $560,000

For each 1$ increase in the ticket, the airline loses 50 passengers, so if we call "x" the increase in the ticket price, we have that the revenue equation will be:

R = (8000-50*x) * (70+x) = 560000 + 8000*x - 50*70*x -50x2

R = -50*x2 + 4500*x + 560000

The value of x that gives the maximum value of a quadratic function is found using the formula:

x = -b/2a

where a and b are coefficients of the quadratic equation (in our case, a = -50 and b = 4500)

So the value of x that gives the maximum revenue is:

x = -4500 / (-100) = 45

Using this value in the revenue equation, we have that the maximum R is:

R = -50*45^2 + 4500*45 + 560000 = $661,250

(To get this revenue, the ticket will cost 70+45 = $115 and there will be 8000-50*45 = 5750 passengers)